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Cool goIITian

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30 Sep 2008 22:42:26 IST

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2^x+3^x+4^x=5^x.Find possible values of x.

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2^x+3^x+4^x=5^x.Find possible values of x.

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Anand

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Joined: 23 Sep 2008

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1 Oct 2008 10:54:03 IST

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d above equation has no solution.

for any value of x, 5^x has d last digit 5.

now, for 2,3 and 4

power 2        3               4              last digit sum

1           2         3               4                  =9

2            4        9             16                =1 9

3           8         27          64                =19

4          16       81       256                 = 13

and above d power 4, d last digit of 2,3 and 4 r periodic

therefore last digit sum will also be periodic.

hence we r getting last digit sum of 2^x+3^x+4^x=9,9,9,3 corresponding to x=1,2,3,4

therefore last digit of L.H.S. is not equal to last digit of R.H.S. for any value of x.

hence above equation has no solution for any value of x.

Hari Shankar

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Joined: 28 Feb 2007

Posts: 2173

1 Oct 2008 11:09:34 IST

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so who said that x has to be a natural number?

venkat v.t

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Joined: 7 Aug 2007

Posts: 244

1 Oct 2008 14:51:23 IST

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no. of solutions 1.

Since the graphs y=2^x+3^x+4^x and y= 5^x intersect at only one point between x=1 and x=2.

Rejaul Makshud

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Joined: 1 Aug 2008

Posts: 96

1 Oct 2008 22:07:04 IST

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2 ^x+ 3 ^x+ 4 ^x= 5 ^x

(2/5) ^x+ (3/5) ^x + (4/5) ^x= 1

Number of solution is one.

It can be solve only graphically. it is not possible to find out exact solution for x.

Simply we draw ths graph of y = (2/5) ^x+ (3/5) ^x + (4/5) ^xand y = 1

so we will get a point of intersection between thses two only one. so the number of solution is one.

ok dear.

From the desk of R. Makshud sir.

Decoder

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Joined: 1 Apr 2007

Posts: 1084

1 Oct 2008 22:36:50 IST

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delete ..2^x..u will have a mathematical answer ...proceed from there..

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